Selfing
F3 1/4 YY + 1/2Yy + 1/4 yy
1 :
2
: 1
All YY
1/3 of the
Yellow F2
3:1 again
2/3 of the
Yellow F2
All yy
Mendel’s First Law
(coined in 1900)
 Law of Equal Segregation:
 
The two alleles of each trait
separate (segregate) during
gamete formation, then unite
at random, one from each
parent, at fertilization.
Conclusions from Monohybrid (single
character segregating) crosses:
Phenotypic Classes Genotypic Classes
yellow:green
3 : 1
YY : Yy : yy
1 : 2 : 1
How could one test this model?
Test Cross
F1
X
Yy
x yy
yellow
green
1/2 Yy
yellow
1/2 yy
green
Test cross allows one to observe the ratio of the alleles
in the F1 parent, because the test cross parent can only
contribute the recessive allele. This way the phenotype
of the plant tells you which allele came from the F1
parent. No need to infer the alleles in the parent from
ratios.
Dihybrid Cross
X
Parents
round yellow X
wrinkled green
F1
round yellow Parents
F1
RR YY
Rr Yy X
rr yy - means the allele type is unknown
F2
round yellow wrinkled yellow
round green wrinkled green
315 101
108
32 R- Yrr YR- yy
rr yy
9 3
3
1 F2
• Wrinkled appearance didn’t stay with green or round with yellow. pollen
Punnett
Square of
Dihybrid
Cross
Each dihybrid plant
produces 4 gamete
types equally
frequently.
eggs
For example, Y can be
with R or r in any
gamete with equal
probability.
Each trait alone = 3:1
Mendel’s Second Law:
Law of independent assortment
Segregation of
alleles of two
different genes
are independent
of one another
Bb
Aa
Bb
aA
B
a
b
A
gametes
B
A
b
a
gametes
Test Cross.
phenotype of test cross progeny directly show the
alleles from the F1 dihybrid.
Rr Yy round yellow 31
F1
Rr Yy
X
x
round yellow
rr Yy wrinkled yellow 27
rr yy
Rr yy round green
wrinkled green
26
rr yy wrinkled green 26
Hybrid Rr Yy Expect 1/4 R Y
in
gametes
1/4 r Y
X
rr yy
r y
test cross progeny
round yellow 31
r y
wrinkled yellow 27
1/4 R y
r y
round green
1/4 r y
r y
wrinkled green 26
Test cross confirms independent assortment of characters
26
Summary
One Characteristic (two phenotypes)
 3:1 F2 phenotypic ratio
 1:1 Test cross phenotypic ratio
Two characteristics (four phenotypes)
•  9 : 3 : 3 : 1 F2 phenotypic ratio
•  1 : 1 : 1 : 1 Test cross phenotypic ratio
Review of Mendelian Genetics
 Law of Equal Segregation. The two alleles of each trait (Y and y)
separate (segregate) during gamete formation, then unite at random,
one from each parent, at fertilization.
 Law of independent assortment. Segregation of alleles of two
different genes are independent of one another.
F2
male gametes
1/2 Y 1/2 y
YY x yy = Yy
P1
P2 F1
female 1/2 Y
gametes
1/2 y
YY
Yy
Yy
yy
Punnett Square
1/4 YY + 1/2Yy + 1/4 yy
1 : 2
: 1
Genotypic ratio 1:2:1
Phenotypic ratio
3:1
Terms
 Genes are what we now call the segregating units that Mendel discovered,
and alleles are the differing forms of a gene that control traits.
 Genotype (the exact alleles present) vs phenotype (observable
consequence) of an organism.
 Homozygous (homozygote) vs heterozygous (heterozygote).
 Mutation = sudden, heritable change in an allele. Most of the allele
variation that we study in genetics arises from mutation.
 Standard allele = wild type = most common allele found in the wild.
 New allele/Variant allele/Mutant allele/Affected allele (depends on the
organism).
Today
 Applying Probability, sampling and combinations(Product rule, sum rule)
to Mendelian genetics
 Extend these ideas to human genetics (pedigrees)
Probability = # times event is expected to happen
# opportunities (trials)
Product Rule (Law of the Product)
The probability of independent
events occurring together is
the product of the
probabilities of the individual
events
Example of tossing two coins.
Punnett Square is a way of depicting the Product Rule.
1/2 Y 1/2 y
1/2 Y YY
Yy
1/2 y Yy
yy
1/4 YY + 1/2Yy + 1/4 yy
1 :
2
: 1
X
yyRR YYrr
YyRr
x
RrYy RrYy
All possible gametes: RY Ry
rY
ry
p(RY gamete) = 1 gamete = 1/4
4 gamete
From monohybrid crosses, chance of the offspring showing
the dominant trait is 3/4 (i. e. 3:1). Using the product rule,
the 9:3:3:1 ratio of a dihybrid cross can be predicted.
Monohybrid.
Yellow to green is
Round to wrinkled is
3/4 to 1/4 (3:1)
3/4 to 1/4 (3:1).
Dihybrid
Yellow round =
Yellow wrinkled =
Green round =
Green wrinkled =
3/4 X 3/4 = 9/16
3/4 X 1/4 = 3/16
1/4 X 3/4 = 3/16
1/4 X 1/4 = 1/16
9
3
3
1
Shorthand way of doing this is (3:1) X (3:1) = 9:3:3:1, helpful when
considering several traits simultaneously.
The probability of either of two
mutually exclusive events occurring
is the sum of their individual
probabilities
• Sum of all the probabilities of all possible
events = 1.
• Probability of something not happening is 1
minus the probability of it happening.
Punnett Square and the Sum Rule.
male gametes
1/2 Y 1/2 y
female
gametes
1/2 Y YY
Yy
1/2 y Yy
yy
Chance of a Yy individual is the sum of the probabilities of
getting the Y from the male and the y from the female (1/2 X
1/2 = 1/4, product rule) and getting the Y from the female and
the y from the male (1/2 X 1/2 = 1/4, product rule). Sum rule 1/4 + 1/4 = 1/2.
What is the probability of
obtaining a round, green seed
from dihybrid (RrYy) cross?
Genotype can be either RRyy or Rryy
Determine the probability that a plant of genotype CcWw
will be produced from the following cross: CcWw x Ccww
1/2 C 1/2 c
Chance of being Cc =
1/2
1/2 C
CC
Cc
1/2 c
Cc
cc
1/2 W 1/2 w
Chance of being Ww = 1/2
1/2 w
Ww
ww
1/2 w
Ww
ww
Chance of being CcWw = 1/2 X 1/2 = 1/4
What fraction of the progeny from
the following cross will have smooth
and red fruit with purple stems?
PpRrAa x ppRRAa
Smooth = Pp = 1/2
P - smooth
p - peach
R - red
r - yellow
A - purple
a - green
Red = Rr or RR = All = 1
Purple = AA (1/4) or Aa (1/2) = 3/4
Smooth, Red, Purple = 1/2 X 1 X 3/4 = 3/8
What is the probability of randomly choosing
three plants that do NOT have smooth and red
fruit with purple stems?
3/8 probability of obtaining smooth, red, purple.
1-3/8 = probability of NOT being smooth red purple = = 5/8
Probability of getting at least one, is 1 minus the probablility
of getting none. Choosing three in a row that are not smooth red purple = 3
(5/8) = 125/512
N!
K! (N-K)!
Binomial expansion
= number of possible ways that K
individuals can be selected from N.
N = Total number of individuals, coins, etc.
K = number of individuals with a specific character.
Example If you have 4 kids, how many ways are there of
having 2 boys and two girls?
boy boy
boy
boy
boy
boy
boy boy
boy
boy
boy boy
4!
2! (4-2)!
=6
=
1X2X3X4
1X2X1X2
4!
=
=
2! 2!
12
2
=6
Example If you have 4 kids, how many ways are there of
having 3 boys and 1 girl?
boy boy boy
boy boy boy
boy boy
boy
boy boy
boy
4!
3! (4-3)!
=
4
1
girl
girl
=4
girl
=
=4
4!
4!
3! 1!
1! (4-1)!
=
=4
girl
4
1
4!
=
1! 3!
=4